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Point A lies at co-ordinate (4,3) and point B at (6,2). Determine: The distance AB

Mathematics

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Here's how to determine the distance AB:

The problem states that Point A is at coordinates $(4,3)$ and Point B is at coordinates $(6,8)$. The student's work correctly calculates the vector from A to B.

  1. Identify the coordinates of the points.

    Point A: (x1,y1)=(4,3)(x_1, y_1) = (4,3)
    Point B: (x2,y2)=(6,8)(x_2, y_2) = (6,8)

  2. Calculate the vector AB⃗\vec{AB} (or B-A).

    The student has already calculated this:

    AB⃗=(x2−x1y2−y1)=(6−48−3)=(25)\vec{AB} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix} = \begin{pmatrix} 6 - 4 \\ 8 - 3 \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \end{pmatrix}

  3. Calculate the distance AB using the magnitude of the vector.

    The distance between two points is the magnitude of the vector connecting them. The magnitude of a vector (xy)\begin{pmatrix} x \\ y \end{pmatrix} is given by x2+y2\sqrt{x^2 + y^2}.

    Distance AB=∥AB⃗∥=(2)2+(5)2\text{Distance AB} = \left\| \vec{AB} \right\| = \sqrt{(2)^2 + (5)^2} Distance AB=4+25\text{Distance AB} = \sqrt{4 + 25} Distance AB=29\text{Distance AB} = \sqrt{29}

    The distance AB is ∗29\boxed*{\sqrt{29}} units.

    2 zimekamilika. Zimebaki 3 leo — tuma inayofuata.

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